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As I see it, the definition of kinetic energy $$T= {1\over2} m u^2 \text { where $u<<c$}$$ comes by using the definition of work $$W= {\int F\cdot\ dx }$$ and we use for the meaning of F(force) the Second Law of Newton: $$F={dp\over dt}=ma$$

Do I understand correctly that the kinetic energy from this point and on becomes a connection link between Newtonian mechanics, theoretical mechanics(Lagrange, Hamilton) and relativity?

If so, this the question: Can there be a definition of energy without the law of Newton?(it seems to me that to use the law of newton is not wrong but strange to the point of view that comes with theoretical mechanics. We remain somehow bounded to a definition of kinetic energy associated with a law that makes the mass something a little more fundamental from let's say charge-by fundamental I mean that something without mass, Newton says it cannot move or have any type of interaction. But if the charge is another way of interaction-via the electromagnetic fields or potentials- shouldn't we search for the possibility of defining the kinetic energy via the charge. What should become of the potentials then? And why mass in the Newton's law?).

Thank you.

PS:if anyone has a recommendation for further study, he can suggest it.

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  • $\begingroup$ the questions in parenthesis at the end are secondary and not needed to be answered without a suggestion on the primary question. $\endgroup$ – Constantine Black Apr 16 '15 at 9:12
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You can define the kinetic energy as you want. As long as they are consistent with each other, definitions are yours to choose.

For example you can take $$T = \frac{1}{2}m v^2 \, ,$$ as a definition and use it to prove Newton's law by requiring energy to be conserved, $$ \frac{dE}{dt} = \frac{dT}{dt} - \frac{dW}{dt} = m v a - F v = 0\, .$$

If you want a less silly answer, you can use the relativistic definition $$ T = \sqrt{p^2 c^2 +m^2 c^4} \, ,$$ ($c$ is the speed of light.) which stays valid for $m=0$.

About your parenthesis: It should not bother you that mass has a special place in the laws of physics as compared to electric charge for example. It is true that in classical physics the mass is the gravitational charge, but is it also (and more importantly) a measure of the inertia of things. Newton's second law, $F = m a$, tells us that it takes more force to accelerate heavy objects. This has nothing to do with gravitational interactions. See this post.

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  • $\begingroup$ Thank you very much for such a fast answer. Allow me only to question this: $\endgroup$ – Constantine Black Apr 16 '15 at 9:24
  • $\begingroup$ Allow me only to question this: The relativistic equation comes by using a lagrangian function invariant under Lorentz,with the demand that in low speeds it becomes T=1/2 m u^2 plus an abstract constant. Also F(for force) has to be understand as an interaction. What leads us to general relativity, as you mention at the post, is of importance. But no matter the kind of interaction, mass, even as a meter of inertia,is fundamental in the interaction- if objects interact via electric field their accellaretion is a function of mass, that is of the quantity participating on gravitational interaction $\endgroup$ – Constantine Black Apr 16 '15 at 9:40
  • $\begingroup$ The whole point about general relativity is that gravitational attraction is indistinguishable from inertia. Gravitational force is only an effective force that we observe because we are not in an inertial frame. I a merry-go-round you feel that are pulled away from the axis of rotation with a force that is proportional to your inertial mass because the source of this force is inertia. GR tells us that Gravity is actually the same. That's why inertial and gravitational mass are the same. $\endgroup$ – Steven Mathey Apr 16 '15 at 9:46
  • $\begingroup$ Okay,thanks. I'll only ask this and won't bother you any more. If the last you wrote is true, then isn't gravity to be understood like something entirely different from electromagnetism or in general the other interactions? Something that maybe is connected more with the space and it's formation because of the presence of what we call mass(this is relativity). $\endgroup$ – Constantine Black Apr 16 '15 at 10:07
  • $\begingroup$ Yes, that's right. If you want to learn about it, I enjoyed this book. $\endgroup$ – Steven Mathey Apr 16 '15 at 10:22

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